If z is complex number such that |z| ≥ 2

Question:- If z is complex number such that |z| ≥ 2 . then the minimum value of |z + 1/2|.

(a) is equal to 5/2

(b) lies int interval (1, 2)

(c) is strictly greater than 5/2

(d) is strictly greater than 3/2 but less than 5/2

Solution:-  |z| = 2

let z = x + iy

|x + iy| = 2

⇒ x² + y² = 4 .  .  . (i)

|z| ≥ 2 is the region on or outside circle whose centre is(0, 0) and radius is  2

The minimum value of |z + 1/2|  is distance of z, which lies on the circle |z| = 2 from (-1/2, 0).

∴  minimum value of |z + 1/2| = Distance between (-1/2, 0) and (-2, 0)

= √(-2 + 1/2)² + 0 = 3/2

Geometrically:- Min |z + 1/2| = AD

Hence, minimum value of |z + 1/2| lies in the interval (1, 2)

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